SME
Aircraft Design
Configuration layout
Use this for “A 30 m² wing (c̄ = 2 m, b = 15 m) with tail arms 6 and 5.5 m. Size the tails for cHT = 0.7 and cVT = 0.06.”
This is an example. Type your own values to start a new problem.
SHT (m²)
=
×
c̄W (m)
×
SW (m²)
LHT (m)
SVT (m²)
=
×
bW (m)
×
SW (m²)
LVT (m)
Formulas
SHT
=
cHT
×
c̄W
×
SW
LHT
7
=
0.7
×
2
×
30
6
SVT
=
cVT
×
bW
×
SW
LVT
4.909
=
0.06
×
15
×
30
5.5
Assumptions
•
The volume coefficients cHT and cVT come from similar airplanes in the course’s table.
•
Each tail arm runs from the wing’s quarter-chord point to the tail’s.
•
The tail’s pull times its arm is what counts, so a longer arm needs a smaller tail.
Step-by-step
WE KNOW
SW = 30 m², c̄W = 2 m, bW = 15 m, LHT = 6 m, LVT = 5.5 m, cHT = 0.7, cVT = 0.06
FIND
horizontal tail area (SHT), vertical tail area (SVT)
Step 1 · Find horizontal tail area (SHT)
Use SHT = cHT × c̄W × SW ÷ LHT
The volume coefficient times the wing’s size, spread over the horizontal tail’s arm.
SHT
=
0.7
×
2
×
30
6
SHT
=
1.4
×
30
6
SHT
=
42
6
SHT = 7 m²
Step 2 · Find vertical tail area (SVT)
Use SVT = cVT × bW × SW ÷ LVT
The volume coefficient times the wing’s size, spread over the vertical tail’s arm.
SVT
=
0.06
×
15
×
30
5.5
SVT
=
0.9
×
30
5.5
SVT
=
27
5.5
SVT = 4.909 m²
Check
Put the numbers back in. Both sides should match.
7
=
0.7
×
2
×
30
6
✓
4.909
=
0.06
×
15
×
30
5.5
✓
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Wing loading for the least thrust
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Mean aerodynamic chord of a tapered wing
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